And as if single-phase circuits aren’t already complicated enough to analyze, three-phase circuits show up. So far in my tutorials for this subject, we’ve been dealing with single-phase circuits. A single-phase AC power system consists of a generator (or the source) connected through a pair of wires (a transmission line) to a load (or the impedance). So if you were to duplicate a single-phase circuit twice, you would have a three-phase circuit. As easy as that. Not really.
A three-phase system is produced by a generator consisting of three sources having the same amplitude (or voltages) and frequency but their phase angles have a difference of 120º. It is considered as by far the most prevalent and most economical polyphase* system, due to the amount of wire required for a three-phase system being less than that required for an equivalent single-phase system. When one-phase or two-phase inputs are required, they are taken from the three-phase system rather than generated independently. That’s why nearly all electric power is generated and distributed in three-phase.
*circuits or systems in which the AC sources operate at the same frequency but different phases.
Here’s how a (a) two-wire type and a (b) three-wire type look like. Note that they are both single-phase systems.
Just because it’s single-phase, doesn’t mean it only has a single loop. A single-phase system means the phase angles are equal, I think. While there’s a single-phase and a three-phase, a two-phase system is also possible.
As you can see, the generator or sources of the two-phase have equal amplitudes, yet having a different value for the phase angle. While the phase angles of a three-phase have a difference of 120º, two-phase have a difference of only 90º.
Here’s a design of a three-phase system:
When a three-phase circuit possess the phase angle difference of its voltage sources but still having its magnitudes equal, it is considered as a balanced three-phase circuit or to contain balanced phase voltages.
A three-phase with balanced phase voltages will look like this in a graph, or whatever this is that displays sinusoids and stuff:
And here is how it is illustrated on a phasor diagram, with the phase sequences: (a) abc or positive sequence, and (b) acb or negative sequence:
So, what about this abc and acb sequence? Well, since the three-phase voltages are 120º out of phase with each other, there are two possible combinations for this. One is the (a) abc or positive sequence phasor diagram where Van leads Vbn, which in turn leads Vcn. This possibility is mathematically expressed as:
Van = Vp∠0º
Vbn = Vp∠-120º
Vcn = Vp∠+120º
Vp is the effective or rms value of the phase voltage. The other possibility is the (b) acb or negative sequence phasor diagram where Van leads Vcn, which in turn leads Vbn. This possibility is given by:
Van = Vp∠0º
Vbn = Vp∠+120º
Vcn = Vp∠-120º
You must have at least wondered how these three balanced voltage sources are connected to each other in a three-phase system, right? No? Me too. There are actually two types of how the three sources (and the three loads) are connected to each other. And those two are called: the wye-connection Y, and the delta-connection Δ.
The one on the left are (a) Y-connected sources, while the other on the right are (b) Δ-connected sources. The three-phase system design that I showed above is an example of a wye-wye connection.
To convert Y-connected sources to Δ-connected sources, or vice versa, you go:
Vl = √3 · Vp ∠(θ + 30º)
Vp = Vl/√3 ∠(θ – 30º)
Vp = |Van| = |Vbn| = |Vcn|
Vl = |Vab| = |Vbc| = |Vca|
It is a common tradition in power systems that the voltage and current, when it comes to three-phase circuits, are always in rms values, unless stated otherwise.
The sources in a three-phase circuit are not the only ones connected in wye or delta, but also the loads.
A set of (a) Y-connected loads and a set of (b) Δ-connected loads. And of course, there is a balanced load, which is considered when the phase impedances are equal in magnitude and in phase. A Y-connected load consists of three impedances connected to a neutral node, while a Δ-connected load consists of three impedances connected around a loop. The load is balanced when the three impedances are equal in either case.
For a balanced Y-connected load,
Z1 = Z2 = Z3 = Zy
Where Zy is the load impedance per phase. While for a balanced Δ-connected load,
Za = Zb = Zc = ZΔ
So, to convert a Y-connected load to a Δ-connected load, or vice versa, you go:
ZΔ = 3Zy
Zy = ZΔ/3
To solve for the current I, just convert the three-phase circuit to a balanced wye-wye connection then get its single-phase equivalent:
Ia = Van/Zy
Ib = Vbn/Zy
Ic = Vcn/Zy
If it is for a balanced delta-delta connection, the formulas would just go:
Ia = Iab – Ica
Ib = Ibc – Iab
Ic = Ica – Ibc
Iab = Vab/ZΔ
Ibc = Vbc/ZΔ
Ica = Vca/ZΔ
And that’s all there is for three-phase circuits. If I have missed some informations or you have some questions about the topic, just comment it down below.